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The inductance `L` of a solenoid of length `l`, whose windings are made of material of density `D` and resistivity `rho`, is (the winding resistance is`R`)A. `(mu_(0))/(4 pi l)(Rm)/(rho D)`B. `(mu_(0))/(4 pi l)(lm)/(rho D)`C. `(mu_(0))/(4 pi l)(R^(2)m)/(rho D)`D. `(mu_(0))/(2 pi R)(l m)/(rho D)` |
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Answer» Correct Answer - A For a sloenoid, `L = mu_(0) N^(2)(A)/(l)`. If `x` is the length of the wire and `a` is the area of cross-section, then `R = (rho x)/(a)` and `m = a xD` `Rm = (rho x)/(a) a x D, x = ((R m)/(rho D))` Also, `x = 2 pi rN, N = (X)/(2pi r) ( :. L = (mu_(0)N^(2)A)/(l))` `:. L = mu_(0)((x)/(2pir))^(2) (pi r^(2))/(l) = (mu_(0))/(4pi l)(Rm)/(rho D)` |
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